The problem is supposed to be a modified version of the lighthouse problem.
Description:
A lighthouse (in point N) is located 4 km away from the shore at sea. A bar is located on the shoreline, 4 km away from point W. Point W is located on the shore such that it is the shortest and the most direct path from the lighthouse to the shore. The shoreline is perpendicular to WN.
We know that a man who is now in the lighthouse wants to go the bar. Also, we know that he rows at the rate of 4 km/h and walks at the rate of 5 km/h. What path should the man choose in order to get to the bar in shortest period of time?
My question is: is the problem as easily solvable as I understand? It seems that no knowledge of calculus is needed to solve it. Or I am missing something?
My solution: we get a triangle which has a perpendicular. We calculate NK, where K is the bar and N is the lighthouse. So now we have 2 paths: NK and NWK. We use the rowing and walking rates to calculate how much time each path will take. Then we compare the results.
As a result, the shortest path is rowing straight from the lighthouse to the bar.
Best Answer
You are correct. The straight line solution is the correct solution. Here is why.
He will row along some straight line path to the shore and land at a point $x$ which is between $0$ and $4$ kilometers from the bar. We want to minimize the total time it takes. The total time will consist of his time on the water and his time on the shore. Making use of the distance formula $d=rt$, where $d$ is distance, $r$ is rate and $t$ is time, and the fact that the total is the sum of its parts, we get
$$t=\frac{\sqrt{16+x^2}}{4}+\frac{4-x}{5}$$ as the total time.
The first term in the sum is the time on the water and the second term represents the time on land.
Taking the derivative with respect to $x$ and setting it to $0$ gives $$\frac{dt}{dx}=\frac{x}{4\sqrt{16+x^2}}-\frac{1}{5}=0\implies x=\frac{4}{5}\sqrt{16+x^2}\implies x^2=\frac{16}{25}(16+x^2)\implies x^2=\frac{256}{9}$$ which gives $x=\frac{16}{3}$
But $\frac{16}{3}\gt 5$ and this expression only works for $x$ between $0$ and $4$.
Moreover, for $x$ between $0$ and $4$, $\frac{dt}{dx}\lt 0$ which means more $x$, less time. Therefore, the minimum time occurs when $x=4$ and the minimum time is $t=\sqrt{2}$ which is what you originally had.